1 norm $||_1$, of non square matrix
$begingroup$
Does $1$ norm exist for non-square matrices? By $1$ norm I mean
$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$
Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.
matrix-norms
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
$endgroup$
add a comment |
$begingroup$
Does $1$ norm exist for non-square matrices? By $1$ norm I mean
$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$
Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.
matrix-norms
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
$endgroup$
$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17
add a comment |
$begingroup$
Does $1$ norm exist for non-square matrices? By $1$ norm I mean
$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$
Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.
matrix-norms
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
$endgroup$
Does $1$ norm exist for non-square matrices? By $1$ norm I mean
$d (x,y)=sum_{i=1}^{n} |x^i-y^i|, x=(x_1,dots, x_n), y=(y_1,dots, y_n)$
Suppose $A$ is $mtimes n, (mne n)$ matrix what can we say about $|A|_1$? Also, can we say $|A|_1=|A^T|_1= |A^TA|_1=|AA^T|_1$? and $|AB|_1le |A|_1|B|_1$ Thanks for helping.
matrix-norms
matrix-norms
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
asked Jan 7 at 10:35
MarkovMarkov
17.3k1059180
17.3k1059180
$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17
add a comment |
$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17
$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$
see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$
This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$
but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$
Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$
Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$
|AA^t|_F=|A^tA|_F
$$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$
(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$
Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$
in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$
see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$
This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$
but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$
Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$
Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$
|AA^t|_F=|A^tA|_F
$$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$
(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$
Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$
in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
$endgroup$
add a comment |
$begingroup$
There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$
see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$
This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$
but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$
Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$
Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$
|AA^t|_F=|A^tA|_F
$$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$
(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$
Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$
in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
$endgroup$
add a comment |
$begingroup$
There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$
see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$
This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$
but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$
Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$
Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$
|AA^t|_F=|A^tA|_F
$$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$
(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$
Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$
in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
$endgroup$
There is no problem to define an "entrywise" matrix norm defined as follows:
$$
|A|_1=|vec(A)|_1=sum_{i,j}|A_{i,j}|
$$
see wikipedia, Matrix norms. It is a norm and the induced distance is (if $A$, $B$ have same dimensions):
$$
d(A,B)=sum_{i,j}|A_{i,j}-B_{i,j}|
$$
This norm is a sub-multiplicative norm (see here):
$$
|AB|_1leq|A|_1|B|_1
$$
but, attention, in general:
$$
|A^tA|_1neq|AA^t|_1
$$
Another example of "entrywise" matrix norm often encountered in practice is the Frobenius norm. This norm is defined as follows:
$$
|A|_F = sqrt{text{tr}(A^tA)}=sqrt{sum_{i,j}|A_{i,j}|^2}
$$
Frobenius norm also fulfills the sub-multiplicative property (Cauchy-Schwarz in action, see
here):
$$
|AB|_Fleq |A|_F|B|_F
$$
Compared to the previous case $|.|_1$, like we have $text{tr}(A^tA)=text{tr}(AA^t)$, Frobenius norm also fulfills the property:
$$
|AA^t|_F=|A^tA|_F
$$
To be complete one must also say a word about Matrix norms induced by vector norms. One can define:
$$
|A|_1=sup_{xneq 0}frac{|Ax|_1}{|x|_1}
$$
(attention despite identical the notation, this norm is different from the previously defined $|vec(.)|_1 $, here we have $|A|_1=max_j sum_i|A_{i,j}|$)
An immediate generalization, valid for any $1leq p leq infty$, is:
$$
|A|_p=sup_{xneq 0}frac{|Ax|_p}{|x|_p}
$$
Such norms are called Matrix norms induced by vector norms, they automatically fulfill the sub-multiplicative property:
$$
|AB|_p leq |A|_p |B|_p
$$
Attention: however in general,
$$
|AA^t|_p neq |A^tA|_p
$$
Also note this important fact, as we are in finite dimension all these previously defined matrix norms are equivalent in the sense that for any two matrix norm $|.|_alpha$ and $|.|_beta$ it exists $r$ and $s$ such that:
$$
forall A, r|A|_alphaleq |A|_beta leq s|A|_alpha
$$
in peculiar if a sequence $nrightarrow (A)_n$ is convergent for a given matrix norm it is also convergent for all the other norms.
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
edited Jan 7 at 12:07
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
answered Jan 7 at 10:45
Picaud VincentPicaud Vincent
1,52439
1,52439
add a comment |
add a comment |
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$begingroup$
What property of the norm are you uncertain about $d$ satisfying?
$endgroup$
– mathworker21
Jan 7 at 10:46
$begingroup$
@OscarLanzi I don't understand what are you talking about, if possible give an example and explain.
$endgroup$
– Markov
Jan 7 at 12:05
$begingroup$
Cannot do so in a comment. If this fails to be clear I am forced to delete.
$endgroup$
– Oscar Lanzi
Jan 7 at 12:17